AP Physics C E&M- 12.3 Magnetic Fields of Current-Carrying Wires and the Biot-Savart Law- Study Notes- New Syllabus
AP Physics C E&M- 12.3 Magnetic Fields of Current-Carrying Wires and the Biot-Savart Law – Study Notes
AP Physics C E&M- 12.3 Magnetic Fields of Current-Carrying Wires and the Biot-Savart Law – Study Notes – per latest Syllabus.
Key Concepts:
- Magnetic Field Produced by a Current-Carrying Wire
- Force on Current-Carrying Wires in a Magnetic Field
Magnetic Field Produced by a Current-Carrying Wire
A moving charge or current produces a magnetic field in the surrounding space. The direction of the magnetic field around a straight current-carrying wire is determined by the right-hand rule: if the thumb points in the direction of current, the curled fingers show the circular magnetic field lines.
Biot–Savart Law (General Expression):
The magnetic field at a point due to a small current element is given by:
\( \delta \vec{B} = \dfrac{\mu_0}{4\pi} \dfrac{I \, \delta \vec{l} \times \hat{r}}{r^2} \)
- \(\mu_0\): permeability of free space \((4\pi \times 10^{-7} \, \mathrm{T \cdot m/A})\).
- \(I\): current in the wire.
- \(\delta \vec{l}\): small element of the wire in the direction of current.
- \(\hat{r}\): unit vector pointing from the current element to the field point.
- \(r\): distance between the current element and the field point.
Magnetic Field Produced by a Current-Carrying Wire
An electric current consists of moving charges, and these moving charges produce a magnetic field around the wire. The strength and direction of this magnetic field depend on the current and the geometry of the wire.
Right-Hand Rule:
- Point the thumb of your right hand in the direction of the conventional current (\(I\)).
- The curl of your fingers shows the circular direction of the magnetic field lines around the wire.
Magnetic Field Around a Long Straight Wire:
Using Ampère’s law or the Biot–Savart law, the magnetic field at a perpendicular distance \(r\) from a long straight wire carrying current \(I\) is:
\( \mathrm{B = \dfrac{\mu_0 I}{2 \pi r}} \)
- \(\mathrm{\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A}\) is the permeability of free space.
- The field decreases with distance (\(1/r\)).
- Field lines are concentric circles around the wire.
Magnetic Field Inside a Current-Carrying Wire:
For a wire of radius \(R\), carrying uniform current distribution:
\( \mathrm{B(r) = \dfrac{\mu_0 I r}{2 \pi R^2}}, \quad (r < R) \)
- Inside the wire, the field increases linearly with distance from the axis.
- At the surface (\(r = R\)), this matches the external formula.
Magnetic Field Due to a Circular Current Loop:
At the center of a circular loop of radius \(R\) carrying current \(I\):
\( \mathrm{B = \dfrac{\mu_0 I}{2R}} \)
- Direction given by the right-hand rule (thumb = current, curl = field).
Properties:
- The field strength increases with current (\(I\)) and decreases with distance (\(r\)).
- Magnetic field lines form closed loops around the current.
- The principle of superposition applies: fields from multiple wires add vectorially.
Example :
A long straight wire carries a current of \(6.0 \, A\). Find the magnetic field at a point \(3.0 \, cm\) from the wire.
▶️ Answer/Explanation
Step 1: Formula: \( \mathrm{B = \dfrac{\mu_0 I}{2 \pi r}} \)
Step 2: Substitute: \( \mathrm{B = \dfrac{(4\pi \times 10^{-7})(6.0)}{2 \pi (0.03)}} \)
Step 3: \( \mathrm{B = 4.0 \times 10^{-5} \, T} \)
Final Answer: The magnetic field is \( \mathrm{4.0 \times 10^{-5} \, T} \), directed circularly around the wire.
Example :
A circular loop of radius \(0.20 \, m\) carries a current of \(3.0 \, A\). Find the magnetic field at the center of the loop.
▶️ Answer/Explanation
Step 1: Formula: \( \mathrm{B = \dfrac{\mu_0 I}{2R}} \)
Step 2: Substitute: \( \mathrm{B = \dfrac{(4\pi \times 10^{-7})(3.0)}{2 (0.20)}} \)
Step 3: \( \mathrm{B = 9.4 \times 10^{-6} \, T} \)
Final Answer: The magnetic field at the center is \( \mathrm{9.4 \times 10^{-6} \, T} \).
Force on Current-Carrying Wires in a Magnetic Field
A wire carrying electric current consists of moving charges (electrons). When placed in a magnetic field, these moving charges experience a magnetic force, which results in a net force on the entire wire.
Mathematical Expression:
The force on a length \(L\) of a straight wire carrying current \(I\) in a magnetic field \(\vec{B}\) is:
\( \mathrm{\vec{F} = I \, \vec{L} \times \vec{B}} \)
- \(\mathrm{I}\) = current in the wire
- \(\mathrm{\vec{L}}\) = length vector of the wire in the direction of current
- \(\mathrm{\vec{B}}\) = magnetic field
Magnitude of the Force:
\( \mathrm{F = I L B \sin \theta} \)
- \(\theta\) = angle between wire and field direction.
- Maximum force when wire is perpendicular (\(\theta = 90^\circ\)).
- No force when wire is parallel to the field (\(\theta = 0^\circ\)).
Direction:
- Right-hand rule: point thumb along the current, fingers along \(\vec{B}\); palm points in the direction of the force.
Applications:
- Electric motors: forces on current-carrying wires produce rotational motion.
- Speakers: wires in magnetic fields vibrate due to alternating current.
Force Between Two Parallel Wires:
- Two parallel wires carrying currents exert magnetic forces on each other.
- Force per unit length between two parallel wires separated by distance \(r\):
\( \mathrm{\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2 \pi r}} \)
- Currents in the same direction → attraction.
- Currents in opposite directions → repulsion.
Example :
A \(0.20 \, m\) long wire carries a current of \(5.0 \, A\) perpendicular to a uniform magnetic field of \(0.30 \, T\). Find the force on the wire.
▶️ Answer/Explanation
Step 1: Formula: \( \mathrm{F = I L B \sin \theta} \).
Step 2: Substitute: \( \mathrm{F = (5.0)(0.20)(0.30)(\sin 90^\circ)} \).
Step 3: \( \mathrm{F = 0.30 \, N} \).
Final Answer: The wire experiences a force of \( \mathrm{0.30 \, N} \).
Example :
Two long parallel wires carry currents of \(I_1 = 8.0 \, A\) and \(I_2 = 6.0 \, A\) in the same direction. If the wires are separated by \(0.05 \, m\), find the force per unit length on each wire.
▶️ Answer/Explanation
Step 1: Formula: \( \mathrm{\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2 \pi r}} \).
Step 2: Substitute: \( \mathrm{\dfrac{F}{L} = \dfrac{(4\pi \times 10^{-7})(8.0)(6.0)}{2 \pi (0.05)}} \).
Step 3: \( \mathrm{\dfrac{F}{L} = 1.92 \times 10^{-4} \, N/m} \).
Step 4: Since currents are in the same direction, the force is attractive.
Final Answer: Each wire experiences an attractive force of \( \mathrm{1.92 \times 10^{-4} \, N/m} \).